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1.Ta có: \(c+ab=\left(a+b+c\right)c+ab\)
\(=ac+bc+c^2+ab\)
\(=a\left(b+c\right)+c\left(b+c\right)\)
\(=\left(b+c\right)\left(a+b\right)\)
CMTT \(a+bc=\left(c+a\right)\left(b+c\right)\)
\(b+ca=\left(b+c\right)\left(a+b\right)\)
Từ đó \(P=\sqrt{\frac{ab}{\left(a+b\right)\left(b+c\right)}}+\sqrt{\frac{bc}{\left(c+a\right)\left(a+b\right)}}+\sqrt{\frac{ca}{\left(b+c\right)\left(a+b\right)}}\)
Ta có: \(\sqrt{\frac{ab}{\left(a+b\right)\left(b+c\right)}}\le\frac{1}{2}\left(\frac{a}{a+b}+\frac{b}{b+c}\right)\)( theo BĐT AM-GM)
CMTT\(\Rightarrow P\le\frac{1}{2}\left(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{a+c}+\frac{b}{a+b}+\frac{c}{b+c}+\frac{a}{a+b}\right)\)
\(\Rightarrow P\le\frac{1}{2}.3\)
\(\Rightarrow P\le\frac{3}{2}\)
Dấu"="xảy ra \(\Leftrightarrow a=b=c\)
Vậy /...
\(\frac{a+1}{b^2+1}=a+1-\frac{ab^2-b^2}{b^2+1}=a+1-\frac{b^2\left(a+1\right)}{b^2+1}\ge a+1-\frac{b^2\left(a+1\right)}{2b}\)
\(=a+1-\frac{b\left(a+1\right)}{2}=a+1-\frac{ab+b}{2}\)
Tương tự rồi cộng lại:
\(RHS\ge a+b+c+3-\frac{ab+bc+ca+a+b+c}{2}\)
\(\ge a+b+c+3-\frac{\frac{\left(a+b+c\right)^2}{3}+a+b+c}{2}=3\)
Dấu "=" xảy ra tại \(a=b=c=1\)
Ta có: \(\sqrt{\frac{ab}{c+ab}}=\sqrt{\frac{ab}{c.1+ab}}=\sqrt{\frac{ab}{c\left(a+b+c\right)+ab}}=\sqrt{\frac{ab}{c\left(b+c\right)+a\left(b+c\right)}}=\sqrt{\frac{ab}{\left(a+c\right)\left(b+c\right)}}\)
\(=\sqrt{\frac{a}{a+c}.\frac{b}{b+c}}\le\frac{1}{2}\left(\frac{a}{a+c}+\frac{b}{b+c}\right)\)( bđt Cosi)
Tương tự như trên: \(\sqrt{\frac{bc}{a+bc}}\le\frac{1}{2}\left(\frac{b}{a+b}+\frac{c}{a+c}\right);\sqrt{\frac{ac}{b+ac}}\le\frac{1}{2}\left(\frac{a}{a+b}+\frac{c}{b+c}\right)\)
=> \(P\le\frac{1}{2}\left(\frac{a}{a+c}+\frac{b}{b+c}+\frac{a}{a+b}+\frac{c}{b+c}+\frac{b}{a+b}+\frac{c}{a+c}\right)=\frac{3}{2}\)
"=" Xảy ra khi và chỉ khi:
\(\frac{a}{a+c}=\frac{b}{b+c}\Leftrightarrow a\left(b+c\right)=b\left(a+c\right)\Leftrightarrow a=b\)
\(\frac{a}{a+b}=\frac{c}{b+c}\Leftrightarrow a=c\)
\(\frac{c}{a+c}=\frac{b}{a+b}\Leftrightarrow b=c\)
\(a+b+c=1\)
Từ các điều trên ta có đc: \(a=b=c=\frac{1}{3}\)
Vậy GTLN của P=3/2 khi và chỉ khi a=b=c=1/3
Ta có: \(\frac{bc}{\sqrt{a+bc}}=\frac{bc}{\sqrt{\left(a+b\right)\left(a+c\right)}}\le\frac{\frac{bc}{a+b}+\frac{bc}{a+c}}{2}\)
\(\frac{ca}{\sqrt{b+ac}}=\frac{ca}{\sqrt{\left(a+b\right)\left(b+c\right)}}\le\frac{\frac{ca}{a+b}+\frac{ca}{b+c}}{2}\)
\(\frac{ab}{\sqrt{c+ab}}=\frac{ab}{\sqrt{\left(a+c\right)\left(b+c\right)}}\le\frac{\frac{ab}{a+c}+\frac{ab}{b+c}}{2}\)
Cộng 3 vế ta được: \(P\le\frac{\frac{bc}{a+b}+\frac{bc}{a+c}+\frac{ca}{a+b}+\frac{ca}{b+c}+\frac{ab}{a+c}+\frac{ab}{b+c}}{2}\)
\(=\frac{\frac{c\left(a+b\right)}{a+b}+\frac{b\left(a+c\right)}{a+c}+\frac{a\left(b+c\right)}{b+c}}{2}=\frac{a+b+c}{2}=\frac{1}{2}\)
Vậy MinP = 1/2
\(\frac{bc}{\sqrt{a+bc}}=\frac{bc}{\sqrt{a.1+bc}}=\frac{bc}{\sqrt{a\left(a+b+c\right)+bc}}=\frac{bc}{\sqrt{\left(a+b\right)\left(a+c\right)}}\le\frac{bc}{a+b}+\frac{bc}{a+c}\)
Bài 4: Áp dụng bất đẳng thức AM - GM, ta có: \(P=\text{}\Sigma_{cyc}a\sqrt{b^3+1}=\Sigma_{cyc}a\sqrt{\left(b+1\right)\left(b^2-b+1\right)}\le\Sigma_{cyc}a.\frac{\left(b+1\right)+\left(b^2-b+1\right)}{2}=\Sigma_{cyc}\frac{ab^2+2a}{2}=\frac{1}{2}\left(ab^2+bc^2+ca^2\right)+3\)Giả sử b là số nằm giữa a và c thì \(\left(b-a\right)\left(b-c\right)\le0\Rightarrow b^2+ac\le ab+bc\)\(\Leftrightarrow ab^2+bc^2+ca^2\le a^2b+abc+bc^2\le a^2b+2abc+bc^2=b\left(a+c\right)^2=b\left(3-b\right)^2\)
Ta sẽ chứng minh: \(b\left(3-b\right)^2\le4\)(*)
Thật vậy: (*)\(\Leftrightarrow\left(b-4\right)\left(b-1\right)^2\le0\)(đúng với mọi \(b\in[0;3]\))
Từ đó suy ra \(\frac{1}{2}\left(ab^2+bc^2+ca^2\right)+3\le\frac{1}{2}.4+3=5\)
Đẳng thức xảy ra khi a = 2; b = 1; c = 0 và các hoán vị
Bài 1: Đặt \(a=xc,b=yc\left(x,y>0\right)\)thì điều kiện giả thiết trở thành \(\left(x+1\right)\left(y+1\right)=4\)
Khi đó \(P=\frac{x}{y+3}+\frac{y}{x+3}+\frac{xy}{x+y}=\frac{x^2+y^2+3\left(x+y\right)}{xy+3\left(x+y\right)+9}+\frac{xy}{x+y}\)\(=\frac{\left(x+y\right)^2+3\left(x+y\right)-2xy}{xy+3\left(x+y\right)+9}+\frac{xy}{x+y}\)
Có: \(\left(x+1\right)\left(y+1\right)=4\Rightarrow xy=3-\left(x+y\right)\)
Đặt \(t=x+y\left(0< t< 3\right)\Rightarrow xy=3-t\le\frac{\left(x+y\right)^2}{4}=\frac{t^2}{4}\Rightarrow t\ge2\)(do t > 0)
Lúc đó \(P=\frac{t^2+3t-2\left(3-t\right)}{3-t+3t+9}+\frac{3-t}{t}=\frac{t}{2}+\frac{3}{t}-\frac{3}{2}\ge2\sqrt{\frac{t}{2}.\frac{3}{t}}-\frac{3}{2}=\sqrt{6}-\frac{3}{2}\)với \(2\le t< 3\)
Vậy \(MinP=\sqrt{6}-\frac{3}{2}\)đạt được khi \(t=\sqrt{6}\)hay (x; y) là nghiệm của hệ \(\hept{\begin{cases}x+y=\sqrt{6}\\xy=3-\sqrt{6}\end{cases}}\)
Ta lại có \(P=\frac{t^2-3t+6}{2t}=\frac{\left(t-2\right)\left(t-3\right)}{2t}+1\le1\)(do \(2\le t< 3\))
Vậy \(MaxP=1\)đạt được khi t = 2 hay x = y = 1
Có bất đẳng thức xy+zt≥x+zy+txy+zt≥x+zy+t với x,z≥0x,z≥0 ,y,t>0y,t>0
Giả sử cc lớn nhất trong các số a,b,ca,b,c thì c≥13c≥13
Do a,b,c≥0a,b,c≥0 nên
Ta có P2≥aa+1+bb+1+cc+1≥a+ba+b+2+cc+1P2≥aa+1+bb+1+cc+1≥a+ba+b+2+cc+1
Mà a+ba+b+2+cc+1−12=1−c3−c+c−12(c+1)=(1−c)(3c−1)(3−c)(2c+2)≥0
Ta có
\(\sqrt{\frac{ab}{c+ab}}=\sqrt{\frac{ab}{c\left(a+b+c\right)+ab}}\)\(=\sqrt{\frac{ab}{\left(c+a\right)\left(c+b\right)}}\)\(=\sqrt{\frac{a}{c+a}}.\sqrt{\frac{b}{c+b}}\)\(\le\frac{1}{2}\left(\frac{a}{c+a}+\frac{b}{c+b}\right)\)
Tương tự, ta có
\(\sqrt{\frac{bc}{a+bc}}\le\frac{1}{2}\left(\frac{b}{a+b}+\frac{c}{a+c}\right)\)
\(\sqrt{\frac{ca}{b+ca}\le\frac{1}{2}\left(\frac{c}{c+b}+\frac{a}{b+a}\right)}\)
Cộng vế theo vế của 3 bđt ta được đpcm
We have \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=3\)
\(\Rightarrow\frac{a+b+c}{abc}=3\Rightarrow a+b+c=3abc\)
Apply inequality Cauchy, we have:
\(\text{Σ}_{cyc}\frac{ab^2}{a+b}\ge3\sqrt[3]{\frac{ab^2}{a+b}.\frac{bc^2}{b+c}.\frac{ca^2}{c+a}}\)
\(=\frac{3abc}{\sqrt[3]{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\ge\frac{a+b+c}{\frac{a+b+b+c+c+a}{3}}=\frac{3}{2}\)
"=" occurs when a = b = c = 1
\(P>=\frac{\left(b\sqrt{a}+c\sqrt{b}+a\sqrt{c}\right)^2}{2\left(a+b+c\right)}\)(bdt svac-xơ)(1)
ta có \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}=3\)
=>\(a+b+c=3abc\)(2)
từ 1 và 2 =>\(P>=\frac{\left(b\sqrt{a}+b\sqrt{c}+a\sqrt{c}\right)^2}{6abc}\)
=>\(P>=\frac{\left(3\sqrt[3]{abc\sqrt{abc}}\right)^2}{6abc}\) (bdt cô si)
=>\(P>=\frac{9abc}{6abc}=\frac{3}{2}\)
xảy ra dấu = khi và chỉ khi a=b=c=1
Ta có: \(ab+bc+ca=abc\)
\(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\)
Đặt: \(A=\frac{a}{bc\left(a+1\right)}+\frac{b}{ca\left(b+1\right)}+\frac{c}{ab\left(c+1\right)}\)
\(\Rightarrow A=\frac{\frac{1}{b}.\frac{1}{c}}{1+\frac{1}{a}}+\frac{\frac{1}{c}.\frac{1}{a}}{1+\frac{1}{b}}+\frac{\frac{1}{b}.\frac{1}{a}}{1+\frac{1}{c}}\)
Đặt: \(\hept{\begin{cases}x=\frac{1}{a}\\y=\frac{1}{b}\\z=\frac{1}{c}\end{cases}}\Rightarrow x+y+z=1\)
\(A=\frac{xy}{z+1}+\frac{yz}{x+1}+\frac{zx}{y+1}\)
Ta có: \(\frac{xy}{z+1}=\frac{xy}{\left(z+x\right)+\left(z+y\right)}\le\frac{1}{4}\left(\frac{xy}{x+z}+\frac{xy}{y+z}\right)\)
Chứng minh tương tự ta được:
\(\frac{yz}{x+1}\le\frac{yz}{x+y}+\frac{yz}{x+z}\)
\(\frac{zx}{y+1}\le\frac{zx}{x+y}+\frac{zx}{y+z}\)
Cộng vế với vế:
\(\Rightarrow A\le\frac{1}{4}\left(x+y+z\right)=\frac{1}{4}\left(đpcm\right)\)
Áp dụng Bđt \(\frac{4}{x+y}\le\frac{1}{x}+\frac{1}{y}\) ta có:
\(\frac{ab}{c+1}=\frac{ab}{\left(a+c\right)+\left(b+c\right)}\le\frac{1}{4}\left(\frac{ab}{a+c}+\frac{ab}{b+c}\right)\)
Tương tự:
\(\frac{bc}{a+1}\le\frac{1}{4}\left(\frac{bc}{b+a}+\frac{bc}{c+a}\right)\)\(;\)\(\frac{ac}{b+1}\le\frac{1}{4}\left(\frac{ac}{a+b}+\frac{ac}{c+b}\right)\)
Cộng theo vế ta được:
\(P\le\frac{1}{4}\left[\left(\frac{ab}{b+c}+\frac{ac}{c+b}\right)+\left(\frac{ab}{a+c}+\frac{bc}{c+a}\right)+\left(\frac{bc}{b+a}+\frac{ac}{a+b}\right)\right]\)
\(=\frac{1}{4}\cdot\left(a+b+c\right)=\frac{1}{4}\)
Dấu = khi \(a=b=c=\frac{1}{3}\)